Bonding and hybridisation in organic compounds Questions in English

(C) In the compound $CH_2 = C = CH_2$ (allene),the central carbon atom is bonded to two other carbon atoms by double bonds.
Since the central carbon atom forms two $\pi$ bonds and two $\sigma$ bonds,it undergoes $sp$ hybridization.
The other options contain only $sp^2$ hybridized carbon atoms.

(D) The structure of $1, 2-$butadiene is $CH_2=C=CH-CH_3$.
In this molecule:
- The first carbon atom $(CH_2=)$ is $sp^2$ hybridized.
- The second carbon atom $(=C=)$ is $sp$ hybridized.
- The third carbon atom $(-CH=)$ is $sp^2$ hybridized.
- The fourth carbon atom $(-CH_3)$ is $sp^3$ hybridized.
Thus,the compound contains $sp, sp^2,$ and $sp^3$ hybridized carbon atoms.

(C) In the $CH_3-C^* \equiv C^*-CH_3$ molecule,both $C^*$ carbon atoms are bonded to one other carbon atom via a single bond and to each other via a triple bond.
Each $C^*$ atom is attached to two atoms in total (one $C$ and one $C$),and there are no lone pairs on the carbon atoms.
The steric number for each $C^*$ atom is $2 + 0 = 2$.
$A$ steric number of $2$ corresponds to $sp$ hybridization.
Therefore,the correct answer is Option $(C)$ $sp$ orbital.

(A) The hybridization of a carbon atom is determined by the number of $\pi$ bonds attached to it.
$1$. The first carbon $(CH_3-)$ is bonded to $3$ hydrogens and $1$ carbon via single bonds ($4$ $\sigma$ bonds),so it is $sp^3$ hybridized.
$2$. The second carbon $(-CH=)$ is bonded to $1$ hydrogen,$1$ single bond,and $1$ double bond ($1$ $\pi$ bond),so it is $sp^2$ hybridized.
$3$. The third carbon $(=C=)$ is bonded to two double bonds ($2$ $\pi$ bonds),so it is $sp$ hybridized.
$4$. The fourth carbon $(=CH_2)$ is bonded to $2$ hydrogens and $1$ double bond ($1$ $\pi$ bond),so it is $sp^2$ hybridized.
Therefore,the hybridization order from left to right is $sp^3, sp^2, sp, sp^2$.

(C) In the molecule $HC\equiv C-CH=CH_2$,let us analyze the hybridization of the carbon atoms involved in the single bond between the second and third carbons.
$1$. The first carbon is $sp$ hybridized due to the triple bond.
$2$. The second carbon is also $sp$ hybridized due to the triple bond.
$3$. The third carbon is $sp^2$ hybridized due to the double bond.
$4$. The fourth carbon is $sp^2$ hybridized due to the double bond.
$5$. The single bond in question is between the second carbon $(sp)$ and the third carbon $(sp^2)$.
Therefore,the hybridization of the carbon atoms in the $C-C$ single bond is $sp-sp^2$.

(C) In the compound $CH_3-CH_2-OH$,the carbon atom marked $C^*$ (the one attached to the hydroxyl group) is bonded to four other atoms (three hydrogens and one oxygen) via single bonds.
Since it forms four sigma bonds and has no lone pairs,its steric number is $4$,which corresponds to $sp^3$ hybridization.
In $H-COOH$,$(NH_2)_2-C=O$,and $CH_3-CHO$,the carbon atom $C^*$ is involved in a double bond with oxygen,making it $sp^2$ hybridized.

(D) To determine the hybridization of the carbon atoms,we number the chain: $C_1H_2 = C_2H - C_3H_2 - C_4H_2 - C_5 \equiv C_6H$.
$C_1$ is $sp^2$ hybridized.
$C_2$ is $sp^2$ hybridized.
$C_3$ is $sp^3$ hybridized.
$C_4$ is $sp^3$ hybridized.
$C_5$ is $sp$ hybridized.
$C_6$ is $sp$ hybridized.
The bond between $C_2$ and $C_3$ involves an $sp^2$ hybridized carbon and an $sp^3$ hybridized carbon.
Therefore,the bond type is $sp^2 - sp^3$.

(B) The hybridization of carbon atoms is determined by the number of sigma bonds and lone pairs attached to them.
$1$. The terminal $CH_3$ groups are bonded to four atoms (three $H$ and one $C$),so they are $sp^3$ hybridized.
$2$. The $CH_2$ group is bonded to four atoms (two $H$ and two $C$),so it is $sp^3$ hybridized.
$3$. The $CH$ groups involved in the double bond are bonded to three atoms (one $H$ and two $C$),so they are $sp^2$ hybridized.
Therefore,the compound contains $sp^2$ and $sp^3$ hybridized carbon atoms.
Structure: $\mathop{CH_3}\limits_{sp^3} - \mathop{CH}\limits_{sp^2} = \mathop{CH}\limits_{sp^2} - \mathop{CH_2}\limits_{sp^3} - \mathop{CH_3}\limits_{sp^3}$

(C) In $(CH_3)_3COH$,the structure is $(CH_3)_3C-OH$.
All the carbon atoms are bonded to four other atoms via single bonds,meaning they are all $sp^3$ hybridized.
In $HCOOH$,the carbonyl carbon is $sp^2$ hybridized.
In $(NH_2)_2CO$,the carbonyl carbon is $sp^2$ hybridized.
In $(CH_3)_3CHO$,the carbonyl carbon is $sp^2$ hybridized.

(C) In the compound $N \equiv C - CH = CH_2$,the carbon atom $(1)$ is bonded to nitrogen by a triple bond,so it is $sp$ hybridised.
The carbon atom $(2)$ is bonded to carbon $(1)$ by a single bond and to the other carbon by a double bond,so it is $sp^2$ hybridised.
Therefore,the hybridisation of carbon $(1)$ and carbon $(2)$ is $sp$ and $sp^2$ respectively.

(A) In the reactant $Br-\overset{1}{CH}=\overset{2}{CH}-Br$,both carbon atoms $1$ and $2$ are bonded to one hydrogen atom,one bromine atom,and one other carbon atom via a double bond,resulting in $sp^2$ hybridisation.
In the product $Br-\overset{3}{CH_2}-\overset{4}{CH_2}-Br$,both carbon atoms $3$ and $4$ are bonded to two hydrogen atoms,one bromine atom,and one other carbon atom via single bonds,resulting in $sp^3$ hybridisation.
Therefore,the correct hybridisation states are $1, 2$ as $sp^2$ and $3, 4$ as $sp^3$.

(D) To determine the hybridization of carbon atoms,we count the number of sigma bonds and lone pairs (if any) attached to each carbon atom.
$(i)$ $CH_3-CH_2-CH_2-CH_3$: All carbon atoms are bonded to four other atoms via single bonds,so all are $sp^3$ hybridized.
$(ii)$ $CH_3-CH=CH-CH_3$: The terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized. Thus,it contains both $sp^2$ and $sp^3$ hybridized carbons.
$(iii)$ $CH_2=CH-CH=CH_2$: All carbon atoms are involved in double bonds,so all are $sp^2$ hybridized.
$(iv)$ $HC\equiv CH$: Both carbon atoms are involved in a triple bond,so both are $sp$ hybridized.
Therefore,only compound $(ii)$ has more than one kind of hybridization for carbon.

(A) To determine if a system is planar,we look at the hybridization of the atoms involved in the structure.
$(i)$ Phenyl group: The benzene ring is planar because all carbon atoms are $sp^2$ hybridized.
$(ii)$ Cyclohexyl group: This is a saturated ring where carbons are $sp^3$ hybridized,adopting a non-planar chair conformation.
$(iii)$ Cyclopentyl group: This is a saturated ring where carbons are $sp^3$ hybridized,adopting a non-planar envelope conformation.
$(iv)$ Butyl group: This is an open-chain alkyl group with $sp^3$ hybridized carbons,which is non-planar.
$(v)$ Ethenyl (vinyl) group: The two carbons are $sp^2$ hybridized,making the system planar.
Therefore,systems $(i)$ and $(v)$ have essentially planar geometry.

(C) The structure of $1, 3$-butadiene is $CH_2=CH-CH=CH_2$.
In this molecule,each carbon atom is bonded to one double bond and two single bonds (or one double and one single bond with hydrogen atoms).
Since each carbon atom is attached to three sigma bonds and zero lone pairs,the steric number is $3$.
Therefore,each carbon atom in $1, 3$-butadiene is $sp^2$ hybridised.

(A) The structure of $1, 2-$butadiene is $CH_2=C=CH-CH_3$.
$1$. The first carbon atom $(CH_2=)$ is bonded to two hydrogens and one double bond,so it is $sp^2$ hybridized.
$2$. The second carbon atom $(=C=)$ is bonded to two double bonds,so it is $sp$ hybridized.
$3$. The third carbon atom $(-CH-)$ is bonded to one hydrogen,one single bond,and one double bond,so it is $sp^2$ hybridized.
$4$. The fourth carbon atom $(-CH_3)$ is bonded to three hydrogens and one single bond,so it is $sp^3$ hybridized.
Thus,the types of hybridization present are $sp, sp^2, \text{ and } sp^3$.

(B) To determine the hybridisation of the underlined carbon atom,we count the number of sigma bonds and lone pairs attached to it.
$(A)$ In $CH_3-\underline{C}OOH$,the underlined carbon is bonded to one oxygen by a double bond and one oxygen by a single bond,making it $sp^2$ hybridized.
$(B)$ In $CH_3-\underline{C}H_2OH$,the underlined carbon is bonded to four atoms (one $C$,two $H$,and one $O$) via single bonds. Since it has $4$ sigma bonds and $0$ lone pairs,it is $sp^3$ hybridized.
$(C)$ In $CH_3-\underline{C}OCH_3$,the underlined carbon is part of a carbonyl group $(C=O)$,making it $sp^2$ hybridized.
$(D)$ In $CH_2=\underline{C}H-CH_3$,the underlined carbon is bonded to one $C$ via a double bond and one $C$ via a single bond,making it $sp^2$ hybridized.
Therefore,the correct option is $B$.

(B) The structure of the molecule is $HC \equiv C - CH = CH_2$.
In this molecule,the $C - C$ single bond is formed between the second carbon (which is $sp$ hybridized due to the triple bond) and the third carbon (which is $sp^2$ hybridized due to the double bond).
Therefore,the hybridisation of the carbons involved in the $C - C$ single bond is $sp - sp^2$.

(C) In the molecule $\mathop {CH_2}\limits^1 = \mathop C\limits^2 = CH_2$,the carbon atom at position $1$ is bonded to two hydrogen atoms and one carbon atom via a double bond. It has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridization.
The carbon atom at position $2$ is bonded to two carbon atoms via two double bonds. It has $2$ sigma bonds and $0$ lone pairs,resulting in $sp$ hybridization.
Therefore,the hybridization of carbon $1$ is $sp^2$ and carbon $2$ is $sp$.

(A) The hybridization of carbon atoms in the given compounds is as follows:
$1$. $\text{Acetonitrile} (CH_3CN)$: The structure is $CH_3-C \equiv N$. The methyl carbon is $sp^3$ and the nitrile carbon is $sp$ hybridized.
$2$. $\text{Acetic acid} (CH_3COOH)$: The structure is $CH_3-C(=O)OH$. The carbonyl carbon is $sp^2$ hybridized.
$3$. $\text{Acetone} (CH_3COCH_3)$: The structure is $CH_3-C(=O)-CH_3$. The carbonyl carbon is $sp^2$ hybridized.
$4$. $\text{Acetamide} (CH_3CONH_2)$: The structure is $CH_3-C(=O)NH_2$. The carbonyl carbon is $sp^2$ hybridized.
Therefore,$\text{Acetonitrile}$ does not contain any $sp^2$ hybridized carbon.

(A) The structure of allyl cyanide is $CH_2=CH-CH_2-C \equiv N$.
To calculate the total number of bonds:
$1$. $CH_2=$ group: $2 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) + $1 \pi$ ($C$=$C$) = $3 \sigma, 1 \pi$.
$2$. $-CH=$ group: $1 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) = $2 \sigma$.
$3$. $-CH_2-$ group: $2 \sigma$ ($C$-$H$) + $1 \sigma$ ($C$-$C$) = $3 \sigma$.
$4$. $-C \equiv N$ group: $1 \sigma$ ($C$-$C$) + $1 \sigma$ ($C$-$N$) + $2 \pi$ ($C$-$N$) = $2 \sigma, 2 \pi$.
Total: $(3+2+3+1) \sigma = 9 \sigma$ bonds and $(1+2) \pi = 3 \pi$ bonds.

(B) The enolic form of acetone is prop$-1-$en$-2-$ol,which has the structure $CH_2=C(OH)-CH_3$.
In this structure,the bonds are as follows:
- $C=C$ bond: $1 \sigma$ and $1 \pi$ bond.
- $C-C$ bond: $1 \sigma$ bond.
- $C-O$ bond: $1 \sigma$ bond.
- $O-H$ bond: $1 \sigma$ bond.
- $C-H$ bonds: $5 \sigma$ bonds (two on terminal $CH_2$ and three on $CH_3$).
Total $\sigma$ bonds = $1+1+1+1+5 = 9$.
Total $\pi$ bonds = $1$.
Oxygen atom has $2$ lone pairs.
Therefore,the correct option is $B$.

(C) The correct answer is $(C)$.
In $CH_{2}=C=CH_{2}$ (Propadiene),the terminal carbon atoms are $sp^{2}$ hybridized because they are bonded to two hydrogen atoms and one carbon atom via a double bond.
The central carbon atom is $sp$ hybridized because it is bonded to two carbon atoms via two double bonds.

(A) In the molecule $H_2C = CH - C \equiv CH$:
$1$. The first carbon atom $(CH_2)$ is bonded to one double bond,so it is $sp^2$ hybridized.
$2$. The second carbon atom $(CH)$ is bonded to one double bond and one single bond,so it is $sp^2$ hybridized.
$3$. The third carbon atom $(C)$ is bonded to one single bond and one triple bond,so it is $sp$ hybridized.
$4$. The fourth carbon atom $(CH)$ is bonded to one triple bond,so it is $sp$ hybridized.
Thus,the hybridization sequence is $sp^2 - sp^2 - sp - sp$.

(A) In the molecule $CH \equiv C - CH = CH_2$,we can label the carbon atoms as follows:
$C^1H \equiv C^2 - C^3H = C^4H_2$
The $C^2$ atom is attached to a triple bond,so it is $sp$ hybridized.
The $C^3$ atom is attached to a double bond,so it is $sp^2$ hybridized.
Therefore,the $C^2 - C^3$ single bond is formed by the overlap of an $sp$ orbital and an $sp^2$ orbital,resulting in $sp - sp^2$ hybridization.

(A) To determine the hybridization of carbon atoms,we look at the number of sigma bonds and lone pairs (if any) attached to each carbon atom:
$(i) CH_3CH_2CH_2CH_3$: All carbon atoms are $sp^3$ hybridized.
$(ii) CH_3-CH=CH-CH_3$: The terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized. Thus,it has more than one type of hybridization.
$(iii) CH_2=CH-C\equiv CH$: The terminal $CH_2$ is $sp^2$,the middle $CH$ is $sp^2$,and the carbons in the triple bond are $sp$ hybridized. Thus,it has more than one type of hybridization.
$(iv) HC\equiv CH$: Both carbon atoms are $sp$ hybridized. Only one type of hybridization is present.
Therefore,compounds $(ii)$ and $(iii)$ contain more than one type of hybridization.

(C) The acidity of a $C-H$ bond depends on the $s$-character of the hybrid orbital of the carbon atom.
Greater $s$-character means the electrons are held more tightly by the nucleus,making the conjugate base more stable.
The $s$-character in different hybridizations is:
$sp$ $(50\%)$ > $sp^2$ $(33.3\%)$ > $sp^3$ $(25\%)$.
Therefore,the $sp-C-H$ bond is the most acidic because the resulting carbanion is the most stable due to the highest electronegativity of the $sp$-hybridized carbon.

(B) Electronegativity of a carbon atom depends on its $s$-character in the hybrid orbital.
Greater the $s$-character,higher is the electronegativity.
$sp$ hybridization has $50\% \ s$-character,$sp^2$ has $33.3\% \ s$-character,and $sp^3$ has $25\% \ s$-character.
Therefore,$sp$-hybridized carbon is the most electronegative.

(B) To determine the hybridization of a carbon atom,we count the number of sigma $(\sigma)$ bonds and lone pairs attached to it. $A$ carbon atom with $4$ sigma bonds is $sp^3$ hybridized.
$A$. In $CH_3\underline{C}OOH$,the underlined carbon is part of a carbonyl group $(C=O)$,so it has $3$ sigma bonds and is $sp^2$ hybridized.
$B$. In $CH_3\underline{C}H_2OH$,the underlined carbon is bonded to one $CH_3$ group,two $H$ atoms,and one $OH$ group. It forms $4$ sigma bonds,so it is $sp^3$ hybridized.
$C$. In $CH_3\underline{C}OCH_3$,the underlined carbon is part of an ether linkage $(C-O-C)$,but the carbon itself is bonded to an oxygen and another carbon. Specifically,in $CH_3-O-CH_3$,the carbons are bonded to $3$ hydrogens and $1$ oxygen,forming $4$ sigma bonds,making them $sp^3$ hybridized. However,looking at the structure $CH_3\underline{C}OCH_3$ (which is likely $CH_3-O-CH_3$),the carbon is $sp^3$. Let's re-evaluate the options. Option $B$ is clearly $sp^3$.
$D$. In $CH_2 = \underline{C}H - CH_3$,the underlined carbon is part of a double bond,so it has $3$ sigma bonds and is $sp^2$ hybridized.
Therefore,the correct option is $B$.

(C) The structure of the compound is $CH \equiv C - CH = CH_2$.
In this molecule,the carbon atoms are hybridized as follows:
$C_1$ (in $CH \equiv$): $sp$
$C_2$ (in $\equiv C-$): $sp$
$C_3$ (in $-CH=$): $sp^2$
$C_4$ (in $=CH_2$): $sp^2$
The $C - C$ single bond is formed between $C_2$ and $C_3$.
Therefore,the hybridization of the carbons involved in the $C - C$ single bond is $sp - sp^2$.

(A) The chemical formula of allyl isocyanide is $CH_2=CH-CH_2-N \equiv C$.
To calculate the number of $\sigma$ and $\pi$ bonds:
$1$. Structure: $H_2C=CH-CH_2-N \equiv C$.
$2$. $\sigma$ bonds: There are $2$ $C-H$ bonds in $CH_2$,$1$ $C-H$ bond in $CH$,$2$ $C-H$ bonds in $CH_2$,$1$ $C-C$ bond,$1$ $C-C$ bond,$1$ $C-N$ bond,and $1$ $C-N$ $\sigma$ bond. Total $\sigma$ bonds = $2+1+2+1+1+1+1 = 9$.
$3$. $\pi$ bonds: There is $1$ $\pi$ bond in the $C=C$ double bond and $2$ $\pi$ bonds in the $N \equiv C$ triple bond. Total $\pi$ bonds = $1+2 = 3$.
Thus,it contains $9\sigma$ and $3\pi$ bonds.

(C) The acidity of a $C-H$ bond depends on the $s$-character of the carbon atom involved in the bond.
As the $s$-character increases,the electronegativity of the carbon atom increases,which stabilizes the resulting carbanion after the removal of a proton.
The $s$-character in different hybridizations is as follows:
$sp^3$ (alkane): $25\% \ s$-character
$sp^2$ (alkene): $33.3\% \ s$-character
$sp$ (alkyne): $50\% \ s$-character
Since the $sp$ hybridized carbon has the highest $s$-character,the $sp C-H$ bond is the most acidic.

(A) The structure of buta-$1,2$-diene is $CH_2=C=CH-CH_3$.
- The first carbon $(C_1)$ is bonded to two hydrogens and one double bond,so it is $sp^2$ hybridized.
- The second carbon $(C_2)$ is bonded to two double bonds,so it is $sp$ hybridized.
- The third carbon $(C_3)$ is bonded to one double bond and one single bond,so it is $sp^2$ hybridized.
- The fourth carbon $(C_4)$ is bonded to four single bonds,so it is $sp^3$ hybridized.
Thus,the hybridization types present are $sp^2, sp, sp^3$.

(B) Let us number the carbon chain as follows: $C_1H_2=C_2H-C_3H_2-C_4 \equiv C_5H$.
$C_1$ is $sp^2$ hybridized.
$C_2$ is $sp^2$ hybridized.
$C_3$ is $sp^3$ hybridized.
$C_4$ is $sp$ hybridized.
$C_5$ is $sp$ hybridized.
The $C_2-C_3$ bond is formed between $C_2$ $(sp^2)$ and $C_3$ $(sp^3)$.
Therefore,the hybridization type of the $C_2-C_3$ bond is $sp^2-sp^3$.

(D) $1$. In $C_4H_{10}$ (butane),all carbon atoms are $sp^3$ hybridized.
$2$. In $CH_3-CH=CH-CH_3$ (but$-2-$ene),the terminal carbons are $sp^3$ hybridized,while the carbons involved in the double bond are $sp^2$ hybridized.
$3$. In $CH_2=CH-C\equiv CH$ (but$-1-$en$-3-$yne),the terminal $CH_2$ carbon is $sp^2$,the middle $CH$ carbon is $sp^2$,and the carbons in the triple bond are $sp$ hybridized.
$4$. Since both $CH_3-CH=CH-CH_3$ and $CH_2=CH-C\equiv CH$ contain carbon atoms with different types of hybridization,the correct answer is $D$.

(C) The structure of propadiene $(CH_2=C=CH_2)$ is as follows:
$1$. The terminal carbon atoms are bonded to two hydrogen atoms and one carbon atom via a double bond,resulting in $sp^2$ hybridization.
$2$. The central carbon atom is bonded to two carbon atoms via two double bonds,resulting in $sp$ hybridization.
Therefore,propadiene contains both $sp$ and $sp^2$ hybridized carbon atoms.

(A) In the reactant $Br(H)C=C(H)Br$,both carbon atoms $1$ and $2$ are involved in a double bond,so they are $sp^2$ hybridized.
In the product $BrCH_2-CH_2Br$,both carbon atoms $3$ and $4$ are involved in only single bonds,so they are $sp^3$ hybridized.
Therefore,$1$ and $2$ are $sp^2$ and $3$ and $4$ are $sp^3$.

(C) First,let us number the carbon chain: $C_1H \equiv C_2 - C_3H_2 - C_4H = C_5H - C_6H_3$.
$C_1$ is $sp$ hybridized (triple bond).
$C_2$ is $sp$ hybridized (triple bond).
$C_3$ is $sp^3$ hybridized (single bonds only).
Therefore,the $C_2 - C_3$ bond is formed by the overlap of an $sp$ hybridized orbital of $C_2$ and an $sp^3$ hybridized orbital of $C_3$.
Thus,the bond type is $sp - sp^3$.

(C) The stability of carbon ions (carbanions) increases with the increase in the $s$-character of the hybrid orbital.
$sp^3$ hybridization has $25\%$ $s$-character.
$sp^2$ hybridization has $33.3\%$ $s$-character.
$sp$ hybridization has $50\%$ $s$-character.
Since higher $s$-character makes the carbon more electronegative,it can better stabilize the negative charge.
Therefore,the increasing order of stability is $sp^3 < sp^2 < sp$.

(D) To determine the hybridization of a carbon atom,we count the number of $\sigma$ bonds and lone pairs attached to it.
$1$. $sp^3$ hybridization: Carbon is bonded to $4$ atoms (all single bonds,$4\sigma$ bonds).
$2$. $sp^2$ hybridization: Carbon is bonded to $3$ atoms (one double bond,$3\sigma$ bonds).
$3$. $sp$ hybridization: Carbon is bonded to $2$ atoms (one triple bond or two double bonds,$2\sigma$ bonds).
In option $D$,the structure is $CH_3-CH_2-C \equiv N$. The underlined carbon atom is bonded to one nitrogen atom via a triple bond and to one carbon atom via a single bond. Thus,it has $2\sigma$ bonds and $0$ lone pairs,resulting in $sp$ hybridization.

(A) The given structure is $CH_2 = CH - C \equiv CH$.
Let us label the carbons as $C_1, C_2, C_3, C_4$ from left to right: $C_1H_2 = C_2H - C_3 \equiv C_4H$.
The single bond is between $C_2$ and $C_3$.
$C_2$ is bonded to one double bond and two single bonds (one with $C_1$,one with $H$,one with $C_3$),so it has $3$ sigma bonds and $0$ lone pairs,resulting in $sp^2$ hybridization.
$C_3$ is bonded to one triple bond and one single bond,so it has $2$ sigma bonds and $0$ lone pairs,resulting in $sp$ hybridization.
Therefore,the hybridization states of the carbon atoms connected by the single bond are $sp^2$ and $sp$.

(A) The keto form of acetone is $CH_3COCH_3$. Its enolic form is $CH_2=C(OH)CH_3$.
In the enolic form $(C_3H_6O)$:
Total number of atoms = $3 (C) + 6 (H) + 1 (O) = 10$.
Number of $\sigma$ bonds = $(\text{Total atoms} - 1) + \text{number of rings} = (10 - 1) + 0 = 9\sigma$ bonds.
Number of $\pi$ bonds = $1$ (in the $C=C$ double bond).
Number of lone pairs = $2$ (on the oxygen atom of the $-OH$ group).

(C) The electronegativity of a carbon atom depends on its hybridization state.
$sp$ hybridized carbon is $50\% \ s$-character,$sp^2$ is $33.3\% \ s$-character,and $sp^3$ is $25\% \ s$-character.
Greater $s$-character implies higher electronegativity.
In option $A$,the starred carbon is $sp^3$ hybridized.
In option $B$,the starred carbon is $sp^2$ hybridized.
In option $C$,the starred carbon is $sp$ hybridized.
In option $D$,the starred carbon is $sp^2$ hybridized.
Since $sp$ hybridization has the highest $s$-character,the carbon in option $C$ is the most electronegative.

(C) The structure of the compound is $CH_3(6)-CH(5)=CH(4)-CH_2(3)-C(2) \equiv CH(1)$.
$1$. Carbon at position $1$ is bonded to one hydrogen and triple-bonded to carbon $2$,so it has $2$ sigma bonds and $0$ lone pairs,resulting in $sp$ hybridization.
$2$. Carbon at position $3$ is bonded to two hydrogens and two carbons,so it has $4$ sigma bonds,resulting in $sp^3$ hybridization.
$3$. Carbon at position $5$ is double-bonded to carbon $4$ and bonded to one hydrogen and one carbon,so it has $3$ sigma bonds,resulting in $sp^2$ hybridization.
Thus,the hybridization for positions $1, 3, 5$ is $sp, sp^3, sp^2$ respectively.

(C) To determine the hybridization of carbon atoms,we look at the number of sigma bonds and lone pairs attached to the carbon atom.
$1$. Acetamide $(CH_3CONH_2)$: The carbonyl carbon $(C=O)$ is $sp^2$ hybridized.
$2$. Acetic acid $(CH_3COOH)$: The carbonyl carbon $(C=O)$ is $sp^2$ hybridized.
$3$. Acetonitrile $(CH_3C \equiv N)$: The methyl carbon is $sp^3$ hybridized,and the nitrile carbon $(C \equiv N)$ is $sp$ hybridized. Thus,it contains no $sp^2$ hybridized carbon.
$4$. Acetone $(CH_3COCH_3)$: The carbonyl carbon $(C=O)$ is $sp^2$ hybridized.
Therefore,the correct option is $C$.

(D) In the given structure,there are $4$ double bonds,each containing one $\pi-$bond.
Therefore,the total number of $\pi-$bonds is $4$.
Since each $\pi-$bond consists of $2$ electrons,the total number of $\pi-$electrons is $4 \times 2 = 8$.

(A) The base strength of a conjugate base is inversely proportional to the acidity of its corresponding acid.
The acidity order of the corresponding hydrocarbons is: $HC \equiv CH > H_2C=CH_2 > H_3C-CH_3$.
This is because the $s$-character in the hybrid orbitals increases from $sp^3$ $(25\%)$ to $sp^2$ $(33.3\%)$ to $sp$ $(50\%)$,making the carbon more electronegative and the conjugate base more stable.
Since the stability of the conjugate base increases in the order $H_3C-CH_2^- < H_2C=CH^- < HC \equiv C^-$,the basicity (which is the inverse of stability) follows the order: $H_3C-CH_2^- > H_2C=CH^- > HC \equiv C^-$.
Therefore,the correct order is $(i) > (ii) > (iii)$.

(B) In the carbanion $CH_{3}-C\equiv C^{-}$,the terminal carbon atom is bonded to another carbon by a triple bond and carries a lone pair of electrons.
The hybridization of a carbon atom involved in a triple bond is $sp$.
Therefore,the lone pair of electrons on the terminal carbon atom is present in an $sp$-hybridized orbital.

(A) $1$. Identify the principal functional group: The carboxylic acid group $(-COOH)$ has the highest priority,so the carbon of the $-COOH$ group is $C_1$.
$2$. Number the chain: $C_1$ (carboxyl carbon),$C_2$ (alpha carbon),$C_3$ (carbon with double bond),$C_4$ (carbon with double bond),$C_5$ (chiral center),$C_6$ (carbon with double bond).
$3$. Analyze $C_3$: The $C_3$ atom is part of a carbon-carbon double bond $(C=C)$,meaning it is bonded to three atoms (one $C$,one $C$,one $H$). Thus,its hybridization is $sp^2$.
$4$. Analyze $C_6$: The $C_6$ atom is part of a terminal carbon-carbon double bond $(CH=CH_2)$,meaning it is bonded to three atoms (one $C$,two $H$). Thus,its hybridization is $sp^2$.
$5$. Conclusion: Both $C_3$ and $C_6$ are $sp^2$ hybridized.

(A) The structure of $\text{hexa-}1, 3-\text{dien-}5-\text{yne}$ is $CH_2=CH-CH=CH-C\equiv CH$.
In this molecule,the carbon atoms involved in double bonds ($sp^2$ hybridized) each have one pure $p-$orbital.
There are $4$ such carbon atoms $(C_1, C_2, C_3, C_4)$,contributing $4$ pure $p-$orbitals.
The carbon atoms involved in the triple bond ($sp$ hybridized) each have two pure $p-$orbitals.
There are $2$ such carbon atoms $(C_5, C_6)$,contributing $2 \times 2 = 4$ pure $p-$orbitals.
Total number of pure $p-$orbitals = $4 + 4 = 8$.

(A) The structure of the molecule $(NC)_3 CCCCHCHBr$ can be expanded as follows:
$N \equiv C-C(C \equiv N)(C \equiv N)-C \equiv C-CH=CH-Br$.
Counting the $\sigma$ bonds:
- $C-N$ bonds: $3$
- $C-C$ bonds: $1$ (between $NC$ and central $C$),$3$ (between central $C$ and $CN$ groups),$2$ (between central $C$ and $C \equiv C$),$1$ (between $C \equiv C$ and $CH$),$1$ (between $CH$ and $CH$),$1$ (between $CH$ and $Br$)
- $C-H$ bonds: $2$
Total $\sigma$ bonds = $3 + 1 + 3 + 2 + 1 + 1 + 1 + 2 = 14$.
Wait,let us re-count carefully:
$N \equiv C-$ ($1$ $\sigma$),$C-C$ ($1$ $\sigma$),$C-C$ ($3$ $\sigma$ to $CN$ groups),$C-C$ ($1$ $\sigma$ to $C \equiv C$),$C-C$ ($1$ $\sigma$ to $CH$),$C-C$ ($1$ $\sigma$ to $CH$),$C-H$ ($2$ $\sigma$),$C-Br$ ($1$ $\sigma$).
Total $\sigma$ bonds = $3$ $(C-N)$ + $1$ ($C-C$ central) + $3$ $(C-CN)$ + $1$ $(C-C \equiv)$ + $1$ $(C \equiv C-C)$ + $1$ $(C-CH)$ + $1$ $(CH-CH)$ + $2$ $(C-H)$ + $1$ $(C-Br)$ = $14$.
Counting $\pi$ bonds:
- $C \equiv N$ groups: $3 \times 2 = 6$
- $C \equiv C$ bond: $2$
- $C=C$ bond: $1$
Total $\pi$ bonds = $6 + 2 + 1 = 9$.
Given the options,there might be a slight variation in interpretation,but based on the structure,the count is $14$ $\sigma$ and $9$ $\pi$. Re-evaluating the options,$13$ and $9$ is the closest match.